The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $2.9$ years; the standard deviation is $0.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living longer than $3.5$ years.
Solution: $2.9$ $2.6$ $3.2$ $2.3$ $3.5$ $2$ $3.8$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $2.9$ years. We know the standard deviation is $0.3$ years, so one standard deviation below the mean is $2.6$ years and one standard deviation above the mean is $3.2$ years. Two standard deviations below the mean is $2.3$ years and two standard deviations above the mean is $3.5$ years. Three standard deviations below the mean is $2$ years and three standard deviations above the mean is $3.8$ years. We are interested in the probability of a lizard living longer than $3.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lizards will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the lizards will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $2.3$ years and the other half $({2.5\%})$ will live longer than $3.5$ years. The probability of a particular lizard living longer than $3.5$ years is ${2.5\%}$.